Question

Help me, please?

Answer 1

A quadratic equation is shown below:

3x2 − 16x + 2 = 0

Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points)

Part B: Solve 9x2 + 3x − 2 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used. (5 points)

Answer 2

I need help finding the radicand.

Answer 3

@peachpi
Do you think you could help me with this one?

Answer 4

i am sure we can do this

Answer 5

Awesome, thank you!

Answer 6

\[3x^2 − 16x + 2 = 0 \] is the equation right?

Answer 7

'Tis correct~

Answer 8

ok
the radicand of \[ax^2+bx+c=0\] is \(b^2-4ac\)

Answer 9

but that does not give you the actual solution
the actual solutions are
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]so lets compute it

Answer 10

actually i see is does not say "find the solutions" it says "Describe the solution(s)"
the radicand in this case is
\[(-6)^2-4\times 3\times 2=232\]

Answer 11

since that is a positive number, but not a perfect square (like say 25 or 16) that means there are two real but irrational solutions

Answer 12

Okay, so we'll have to find those irrational solutions, right? But how?

Answer 13

they do not ask you to find them, but we can

Answer 14

solution is
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] lets plug in the numbers and see what we get

Answer 15

we already know \(b^2-4ac=232\)

Answer 16

plugging them in give
\[\frac{16\pm\sqrt{232}}{2\times 3}\]

Answer 17

then some more arithmetic is needed

Answer 18

Okay, so for the top part does that mean you could add or subtract or is there a specific one you're supposed to use?

Answer 19

it is a short hand for saying there are two different solution
one with the plus, and another with the minus

Answer 20

Oh alright, that makes sense. But how would you solve the top part? We've already established that the 232 ends up two real irrational numbers, if we're trying to figure those out now, what do we do on the top?

Answer 21

we can to that but i want to stress that the question does not ask for the actual solutions, only to "describe " them
the description is two real irrational solutions
now lets find them

Answer 22

Right, okay.

Answer 23

first of \[232=4\times 58\] so
\[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]

Answer 24

therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]

Answer 25

now cancel a 2 top and bottom to finish with
\[\frac{8\pm\sqrt{58}}{3}\]

Answer 26

since 58 is not a prefect square, \(\sqrt{58}\) is irrational

Answer 27

Oh okay, I see that now.

Answer 28

that is your final answer to the first one
second one is easier

Answer 29

Hopefully, because I'm terrible at those cx

Answer 30

the second one factors

Answer 31

Oh so its just a factoring problem?

Answer 32

yeah pretty much

Answer 33

\[ 9x^2 + 3x − 2 = 0 \\
(3x+1)(3x-2)=0\] set each factor equal to zero and solve for \(x\)

Answer 34

Is that the finished product?

Answer 35

no that is just what you get when you factor
to finish solving for \(x\) solve\[3x+1=0\] and also \[3x-2=0\]

Answer 36

The first one would be
\[-\frac{ 1 }{ 3 }\]
Right?

Answer 37

And the second one
\[\frac{ 2 }{ 3 }\]

Answer 38

yup

Answer 39

oh wait a second

Answer 40

dang i made a mistake sorry

Answer 41

Its alright ^.^

Answer 42

it is not \[(3x+1)(3x-2)=0\] it is \[(3x-1)(3x+2)=0\]

Answer 43

so solutions are \(\frac{1}{3}\) and \(-\frac{2}{3}\)

Answer 44

Alright cool, so that's it?

Answer 45

yup thats it

Answer 46

Thank you so much! That helped so much!

Answer 47

yw